Question Number 119634 by Ar Brandon last updated on 25/Oct/20
![Q1 If f:R→R is defined by f(x)=[x]+[x+(1/2)]+[x+(2/3)]−3x+5 where [x] is the integral part of x, then a period of f is (A) 1 (B) 2/3 (C) 1/2 (D) 1/3 Q2 Let a<c<b such that c−a=b−c. If f:R→R is a function satisfying the relation f(x+a)+f(x+b)=f(x+c) for all x∈R then a period of f is (A) (b−a) (B) 2(b−a) (C) 3(b−a) (D) 4(b−a)](https://www.tinkutara.com/question/Q119634.png)
$$\mathrm{Q1} \\ $$$$\mathrm{If}\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{by} \\ $$$$\:\:\:\:\:\:\:\:{f}\left(\mathrm{x}\right)=\left[\mathrm{x}\right]+\left[\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right]+\left[\mathrm{x}+\frac{\mathrm{2}}{\mathrm{3}}\right]−\mathrm{3x}+\mathrm{5} \\ $$$$\mathrm{where}\:\left[\mathrm{x}\right]\:\mathrm{is}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{part}\:\mathrm{of}\:\mathrm{x},\:\mathrm{then}\:\mathrm{a}\:\mathrm{period}\:\mathrm{of}\:{f}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{2}/\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{C}\right)\:\mathrm{1}/\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{1}/\mathrm{3} \\ $$$$ \\ $$$$\mathrm{Q2} \\ $$$$\mathrm{Let}\:{a}<\mathrm{c}<\mathrm{b}\:\mathrm{such}\:\mathrm{that}\:\mathrm{c}−{a}=\mathrm{b}−\mathrm{c}.\:\mathrm{If}\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{function}\:\mathrm{satisfying}\:\mathrm{the}\:\mathrm{relation} \\ $$$$\:\:{f}\left(\mathrm{x}+{a}\right)+{f}\left(\mathrm{x}+\mathrm{b}\right)={f}\left(\mathrm{x}+\mathrm{c}\right)\:\:\mathrm{for}\:\mathrm{all}\:\mathrm{x}\in\mathbb{R} \\ $$$$\mathrm{then}\:\mathrm{a}\:\mathrm{period}\:\mathrm{of}\:{f}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\left(\mathrm{b}−{a}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{2}\left(\mathrm{b}−{a}\right) \\ $$$$\left(\mathrm{C}\right)\:\mathrm{3}\left(\mathrm{b}−{a}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{4}\left(\mathrm{b}−{a}\right) \\ $$
Answered by Olaf last updated on 26/Oct/20
![Q2. c−a = b−c c = ((a+b)/2) f(x+c) = f(x+a)+f(x+b) f(x) = f(x+a−c)+f(x+b−c) f(x) = f(x+a−((a+b)/2))+f(x+b−((a+b)/2)) f(x) = f(x−(1/2)(b−a))+f(x+(1/2)(b−a)) f(x−(1/2)(b−a)) = f(x−(b−a))+f(x) f(x)−f(x+(1/2)(b−a)) = f(x−(b−a))+f(x) f(x+(1/2)(b−a)) = −f(x−(b−a)) f(x+(3/2)(b−a)) = −f(x) f(x+3(b−a)) = −f(x+(3/2)(b−a)) f(x+3(b−a)) = −[−f(x)] = f(x) Period is 3(b−a)](https://www.tinkutara.com/question/Q119671.png)
$$ \\ $$$$\mathrm{Q2}. \\ $$$${c}−{a}\:=\:{b}−{c} \\ $$$${c}\:=\:\frac{{a}+{b}}{\mathrm{2}} \\ $$$$ \\ $$$${f}\left({x}+{c}\right)\:=\:{f}\left({x}+{a}\right)+{f}\left({x}+{b}\right) \\ $$$${f}\left({x}\right)\:=\:{f}\left({x}+{a}−{c}\right)+{f}\left({x}+{b}−{c}\right) \\ $$$${f}\left({x}\right)\:=\:{f}\left({x}+{a}−\frac{{a}+{b}}{\mathrm{2}}\right)+{f}\left({x}+{b}−\frac{{a}+{b}}{\mathrm{2}}\right) \\ $$$${f}\left({x}\right)\:=\:{f}\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\left({b}−{a}\right)\right)+{f}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\left({b}−{a}\right)\right) \\ $$$$ \\ $$$${f}\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\left({b}−{a}\right)\right)\:=\:{f}\left({x}−\left({b}−{a}\right)\right)+{f}\left({x}\right) \\ $$$${f}\left({x}\right)−{f}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\left({b}−{a}\right)\right)\:=\:{f}\left({x}−\left({b}−{a}\right)\right)+{f}\left({x}\right) \\ $$$${f}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\left({b}−{a}\right)\right)\:=\:−{f}\left({x}−\left({b}−{a}\right)\right) \\ $$$$ \\ $$$${f}\left({x}+\frac{\mathrm{3}}{\mathrm{2}}\left({b}−{a}\right)\right)\:=\:−{f}\left({x}\right) \\ $$$$ \\ $$$$\:{f}\left({x}+\mathrm{3}\left({b}−{a}\right)\right)\:=\:−{f}\left({x}+\frac{\mathrm{3}}{\mathrm{2}}\left({b}−{a}\right)\right) \\ $$$$\:{f}\left({x}+\mathrm{3}\left({b}−{a}\right)\right)\:=\:−\left[−{f}\left({x}\right)\right]\:=\:{f}\left({x}\right) \\ $$$$ \\ $$$$\mathrm{Period}\:\mathrm{is}\:\mathrm{3}\left({b}−{a}\right) \\ $$
Answered by Olaf last updated on 26/Oct/20
![Q1. f(x) = [x]+[x+(1/2)]+[x+(2/3)]−3x+5 f(x+1) = [x+1]+[(x+(1/2))+1] +[(x+(2/3))+1]−3(x+1)+5 f(x+1) = [x]+1+[x+(1/2)]+1 +[x+(2/3)]+1−3(x+1)+5 f(x+1) = [x]+[x+(1/2)]+[x+(2/3)]−3x+5 f(x+1) = f(x) f is clearly at least 1−periodic. f(0) = [0]+[(1/2)]+[(2/3)]−3(0)+5 = 5 f((1/3)) = [(1/3)]+[(5/6)]+[1]−3((1/3))+5 = 5 f((1/3)) = f(0) By extension, in the general case : f(x+(1/3)) = f(x) f is (1/3)−periodic.](https://www.tinkutara.com/question/Q119675.png)
$$\mathrm{Q1}. \\ $$$${f}\left({x}\right)\:=\:\left[{x}\right]+\left[{x}+\frac{\mathrm{1}}{\mathrm{2}}\right]+\left[{x}+\frac{\mathrm{2}}{\mathrm{3}}\right]−\mathrm{3}{x}+\mathrm{5} \\ $$$${f}\left({x}+\mathrm{1}\right)\:=\:\left[{x}+\mathrm{1}\right]+\left[\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{1}\right] \\ $$$$+\left[\left({x}+\frac{\mathrm{2}}{\mathrm{3}}\right)+\mathrm{1}\right]−\mathrm{3}\left({x}+\mathrm{1}\right)+\mathrm{5} \\ $$$${f}\left({x}+\mathrm{1}\right)\:=\:\left[{x}\right]+\mathrm{1}+\left[{x}+\frac{\mathrm{1}}{\mathrm{2}}\right]+\mathrm{1} \\ $$$$+\left[{x}+\frac{\mathrm{2}}{\mathrm{3}}\right]+\mathrm{1}−\mathrm{3}\left({x}+\mathrm{1}\right)+\mathrm{5} \\ $$$${f}\left({x}+\mathrm{1}\right)\:=\:\left[{x}\right]+\left[{x}+\frac{\mathrm{1}}{\mathrm{2}}\right]+\left[{x}+\frac{\mathrm{2}}{\mathrm{3}}\right]−\mathrm{3}{x}+\mathrm{5} \\ $$$${f}\left({x}+\mathrm{1}\right)\:=\:{f}\left({x}\right) \\ $$$${f}\:\mathrm{is}\:\mathrm{clearly}\:\mathrm{at}\:\mathrm{least}\:\mathrm{1}−\mathrm{periodic}. \\ $$$$ \\ $$$${f}\left(\mathrm{0}\right)\:=\:\left[\mathrm{0}\right]+\left[\frac{\mathrm{1}}{\mathrm{2}}\right]+\left[\frac{\mathrm{2}}{\mathrm{3}}\right]−\mathrm{3}\left(\mathrm{0}\right)+\mathrm{5}\:=\:\mathrm{5} \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:=\:\left[\frac{\mathrm{1}}{\mathrm{3}}\right]+\left[\frac{\mathrm{5}}{\mathrm{6}}\right]+\left[\mathrm{1}\right]−\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\mathrm{5}\:=\:\mathrm{5} \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:=\:{f}\left(\mathrm{0}\right) \\ $$$$\mathrm{By}\:\mathrm{extension},\:\mathrm{in}\:\mathrm{the}\:\mathrm{general}\:\mathrm{case}\:: \\ $$$${f}\left({x}+\frac{\mathrm{1}}{\mathrm{3}}\right)\:=\:{f}\left({x}\right) \\ $$$${f}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{periodic}. \\ $$
Commented by Ar Brandon last updated on 26/Oct/20
Thank you so very much Sir ��